Example Problem Post

This problem is from the Columbus State University problem of the week. Link

Problem

Given a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5, find the value of a + b + c + d.

Solution

Note that this string of equalities gives us the following relations.

  1. a + 1 = b + 2
  2. a + 1 = c + 3
  3. a + 1= d + 4
  4. a + 1 = a + b + c + d + 5

We will first solve for a which will in turn allow us to solve for the other values. Equations (1), (2), and (3) tell us that b = a - 1, c = a -2, and d = a - 3, respectively. If we substitute these values into equation (4) we find

a + 1 = a + (a- 1) + (a-2) + (a-3) + 5

2 = 3a

a = 2/3

Thus we can use equations (1), (2), and (3) to find that b = -1/3, c = -4/3, and d = -7/3. Their sum is

a + b + c + d = 2/3 - 1/3 - 4/3 - 7/3 = -10/3

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One response to “Example Problem Post

  1. Another way to solve this problem is using linear algebra. Note that the string of equalities gives us the following relations:
    -4 = b + c + d
    -3 = a + c + d
    -2 = a + b + d
    -1 = a + b + c
    This allows us to set up two matrices:
    A = \left[\begin{array}{cccc}0 & 1 & 1 & 1\\1 & 0 & 1 & 1\\1& 1& 0 & 1\\ 1& 1& 1& 0 \end{array}\right]
    and
    b = \left[\begin{array}{c}-4 \\ -3 \\ -2 \\ -1\end{array}\right]
    Then when you compute
    A^{-1}b = \left[\begin{array}{c}2/3 \\ -1/3 \\ -4/3 \\ -7/3\end{array}\right]
    you find a = 2/3 , b = -1/3, c = -4/3, d = -7/3.

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